Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $t = \dfrac{4q - 32}{-5q^2 - 20q + 60} \times \dfrac{q^2 + 14q + 48}{3q + 24} $
Answer: First factor out any common factors. $t = \dfrac{4(q - 8)}{-5(q^2 + 4q - 12)} \times \dfrac{q^2 + 14q + 48}{3(q + 8)} $ Then factor the quadratic expressions. $t = \dfrac {4(q - 8)} {-5(q + 6)(q - 2)} \times \dfrac {(q + 6)(q + 8)} {3(q + 8)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {4(q - 8) \times (q + 6)(q + 8) } { -5(q + 6)(q - 2) \times 3(q + 8)} $ $t = \dfrac {4(q + 6)(q + 8)(q - 8)} {-15(q + 6)(q - 2)(q + 8)} $ Notice that $(q + 6)$ and $(q + 8)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {4\cancel{(q + 6)}(q + 8)(q - 8)} {-15\cancel{(q + 6)}(q - 2)(q + 8)} $ We are dividing by $q + 6$ , so $q + 6 \neq 0$ Therefore, $q \neq -6$ $t = \dfrac {4\cancel{(q + 6)}\cancel{(q + 8)}(q - 8)} {-15\cancel{(q + 6)}(q - 2)\cancel{(q + 8)}} $ We are dividing by $q + 8$ , so $q + 8 \neq 0$ Therefore, $q \neq -8$ $t = \dfrac {4(q - 8)} {-15(q - 2)} $ $ t = \dfrac{-4(q - 8)}{15(q - 2)}; q \neq -6; q \neq -8 $